Question

A projectile is thrown with a velocity of $10\sqrt{2}m{s}^{-1}$ at an angle of ${45}^o$ with horizontal. The time interval between the moments when its speed is $\sqrt{125}m{s}^{-1}$ is
$g=10m{s}^{-2}$

• A. $1.0s$
• B. $0.6s$
• C. $1.2s$
• D. $0.8s$

Answer 1

The correct option is A $1.0s$

$v$ is the velocity of the particle where the speed is $\sqrt{125}m{s}^{-1}$
As the horizontal velocity is constant, $u\cos \theta =v\cos \psi$
$10\sqrt{2}\times \frac{1}{\sqrt{2}}=\sqrt{125}\cos \psi$
$\cos \psi =\frac{2}{\sqrt{5}}$
$\Rightarrow \sin \psi =\sqrt{1-(\frac{2}{5}{)}^2}=\frac{1}{\sqrt{5}}$
Between points $A$ and $B$
$t=\frac{2v\sin \psi }{g}=1s$